Combinatorics - Math :)

aloha · 2010-12-02 23:55:11

Hi, I need to solve this math problem. I'm in serious troubles , so I want to ask, that some body can't help me. Thx a lot!

We have dishonest six-sided cube. We know that number 6 drop three times more than number 1, number 5 drops two times more that number 2 and number 4 has same drop like number 3.

a) How many cubes like that exist? Describe, how they look like. (probability of each side)
b) What is the average value of dots in one throw. (result isn't number, but expression)
c) What is the biggest average value of dots in one throw, which we can get at the appropriate constructions?

.....excuse my English

DrSteve · 2010-12-03 00:13:36

Here's something to start you off:

P(6)=3P(1), P(5)=2P(2), P(4)=P(3). Let x = P(1), y = P(2), z = P(3). Since probabilities add to 1, we have

x + y + z + z + 2y + 3x = 1, equivalently 4x + 3y + 2z = 1

majkld · 2010-12-03 01:28:01

DrSteve wrote:

Here's something to start you off:
P(6)=3P(1), P(5)=2P(2), P(4)=P(3). Let x = P(1), y = P(2), z = P(3). Since probabilities add to 1, we have
x + y + z + z + 2y + 3x = 1, equivalently 4x + 3y + 2z = 1

yes I have the same way of thinking. Now there is the question of amount of cubes...

and i b) part, I have idea, that i will express x, y and z and i will put this into formula Ex = 1*x + 2*y + 3*z + 4*z + 5*2y + 6*3x. Is my idea correct? thx for respond.

aloha · 2010-12-03 02:00:02

Great
but it has not calculated the one?? I really have not found any competent mathematician? Thanks

Last edited by aloha (2010-12-03 02:00:38)

soroban · 2010-12-03 06:37:40

. .

.

aloha · 2010-12-03 07:21:23

Thank you for your help
Then I will resolve the last example c) ....Anybody solve it please? Thanks a lot

Last edited by aloha (2010-12-03 07:22:08)

bobbym · 2010-12-03 07:23:04

Hi aloha;

In your own words what is left that you do not understand? Please post it.

aloha · 2010-12-03 08:05:56

Hi bobbym

I would like to formulate my question better, but my English is very bad! I am using Google Translate

My question on this part of the example:
c) What is the biggest average value of dots in one throw, which we can get at the appropriate constructions?

.....excuse my English

Last edited by aloha (2010-12-03 08:06:44)

bobbym · 2010-12-03 08:29:39

Hi Aloha;

In order to do this well I would appreciate more information about the die.
I figure that no face has a probability of 0. With that constraint and the others I figure it to be very close to 4.75

The empirical result is as close to 4.75 as you want it to be. Mathematically you can establish 19 / 4 as the maximum average value only if some faces have 0 probability.

aloha · 2010-12-03 08:47:27

The biggest average value is due out 4 and 3/4.... 4.75 really!! and could you show me how you came to this result? Using a formula or what? Thanks a lot

.....excuse my English

Last edited by aloha (2010-12-03 08:51:26)

bobbym · 2010-12-03 08:58:49

Hi aloha;

Please reread my earlier post. Mathematically you can establish 19 / 4 as the maximum average value only if some faces have 0 probability! 19 / 4 is the maximum when x = 1 / 4 , y = 0, z = 0. That means 4 faces never come up!

This is verified by using the method of Lagrangian Multipliers and Mathematica. To get a real answer ( one that all faces can come up ) you need more conditions on the faces of your die.

aloha · 2010-12-03 21:32:41

so once again an example c) :
What is the maximum average value of eyes one can get with a single roll, which can be reached considering proper construction of the die?

Last edited by aloha (2010-12-03 21:32:59)

bobbym · 2010-12-03 23:02:02

Hi aloha;

If you mean an ordinary die. 3.5 is the expected value of that die. That is also the maximum.

Remember, as I told you for your die the maximum occurs at x = 1 / 4 , y = 0, z = 0. It is 4.75.
This answer is paradoxical it means that 4 faces out of 6 have zero probability. You would need a few more conditions on the faces to get a realistic answer.

aloha · 2010-12-04 00:13:02

bobbym, soroban, DrSteve: thank you very much

Last edited by aloha (2010-12-04 00:13:34)

bobbym · 2010-12-04 00:25:25

Hi aloha;

Your welcome!

If you are satisfied that only 1 and 6 can come up then c has indeed been answered. Remember to achieve the maximum 4.75, x = 1/4, y,z = 0. This means that (2,3,4,5) have a probability of 0. Is this a school problem or one you made up?

majkld2 · 2010-12-04 03:54:06

bobbym wrote:

Hi aloha;
Your welcome!
If you are satisfied that only 1 and 6 can come up then c has indeed been answered. Remember to achieve the maximum 4.75, x = 1/4, y,z = 0. This means that (2,3,4,5) have a probability of 0. Is this a school problem or one you made up?

thx too. And yes that's school problem :-)

majkd2 · 2010-12-04 04:28:50

I woud like to ask something in part a). How many dice like that can exist. I think it is infinity but I'm not sure. Thx for responds.

bobbym · 2010-12-04 10:02:06

Hi majkld2;

majkld2 wrote:

thx too. And yes that's school problem :-)

Welcome to the forum. May I ask how you know that is a school problem?

majkd2 wrote:

I woud like to ask something in part a). How many dice like that can exist. I think it is infinity but I'm not sure. Thx for responds.

In reality what you are asking is how many solutions are there to the following equation?

4x + 3y + 2z = 1 with 0 ≤ x,y,z ≤ 1

The answer is there is an infinite number of them, therefore an infinite number of die like that.

aloha · 2010-12-04 21:57:46

Hi bobbym;
May I ask why it works infinity? Could you please bring something more? Thanks

bobbym · 2010-12-04 22:04:45

Hi aloha;

There is an infinite number of dice like that because 4x + 3y + 2z = 1 with 0 ≤ x ≤ 1 / 4, 0 ≤ y ≤ 1 / 3, 0 ≤ z ≤ 1 / 2 has an infinite number of solutions.

aloha · 2010-12-05 20:46:33

Hi bobbym;

Thank you a lot!!

bobbym · 2010-12-05 20:51:16

Hi aloha;

Your welcome. I enjoyed the problem.

Math Is Fun Forum

#1 2010-12-02 23:55:11

Combinatorics - Math :)

#2 2010-12-03 00:13:36

Re: Combinatorics - Math :)

#3 2010-12-03 01:28:01

Re: Combinatorics - Math :)

#4 2010-12-03 02:00:02

Re: Combinatorics - Math :)

#5 2010-12-03 06:37:40

Re: Combinatorics - Math :)

#6 2010-12-03 07:21:23

Re: Combinatorics - Math :)

#7 2010-12-03 07:23:04

Re: Combinatorics - Math :)

#8 2010-12-03 08:05:56

Re: Combinatorics - Math :)

#9 2010-12-03 08:29:39

Re: Combinatorics - Math :)

#10 2010-12-03 08:47:27

Re: Combinatorics - Math :)

#11 2010-12-03 08:58:49

Re: Combinatorics - Math :)

#12 2010-12-03 21:32:41

Re: Combinatorics - Math :)

#13 2010-12-03 23:02:02

Re: Combinatorics - Math :)

#14 2010-12-04 00:13:02

Re: Combinatorics - Math :)

#15 2010-12-04 00:25:25

Re: Combinatorics - Math :)

#16 2010-12-04 03:54:06

Re: Combinatorics - Math :)

#17 2010-12-04 04:28:50

Re: Combinatorics - Math :)

#18 2010-12-04 10:02:06

Re: Combinatorics - Math :)

#19 2010-12-04 21:57:46

Re: Combinatorics - Math :)

#20 2010-12-04 22:04:45

Re: Combinatorics - Math :)

#21 2010-12-05 20:46:33

Re: Combinatorics - Math :)

#22 2010-12-05 20:51:16

Re: Combinatorics - Math :)

Board footer